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3 years ago

A 860 kΩ resistor has 34 μA of current. What is the supply voltage for this electric circuit?

Engineering

2 answers:

Juliette [100K]3 years ago

8 0

Explanation:

first changing kilo ohm to ohm

860000 = 860 kΩ

and change 34 micro ampare to ampare

34 μA=3.4×10^-5

recalling the equation V=I*R

V= 3.4×10^-5×860000

v=29.24

deff fn [24]3 years ago

6 0

Answer:

v=IR

V= 860 x 10^3 x 34 x 10^-6

V=29240 x 10^-3

V=29.24 V

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2 years ago

A square-thread power screw is used to raise or lower the basketball board in a gym, the weight of which is W = 100kg. See the f

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Answer:

power = 49.95 W

and it is self locking screw

Explanation:

given data

weight W = 100 kg = 1000 N

diameter d = 20mm

pitch p = 2mm

friction coefficient of steel f = 0.1

Gravity constant is g = 10 N/kg

solution

we know T is

T = w tan(α + φ ) $\frac{dm}{2}$ ...................1

here dm is = do - 0.5 P

dm = 20 - 1

dm = 19 mm

and

tan(α) = $\frac{L}{\pi dm}$ ...............2

here lead L = n × p

so tan(α) = $\frac{2\times 2}{\pi 19}$

α = 3.83°

and

f = 0.1

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so that φ = 5.71°

and now we will put all value in equation 1 we get

T = 1000 × tan(3.83 + 5.71 ) $\frac{19\times 10^{-3}}{2}$

T = 1.59 Nm

power = $\frac{2\pi N \ T }{60}$ .................3

put here value

power = $\frac{2\pi \times 300\times 1.59}{60}$

power = 49.95 W

and

as φ > α

so it is self locking screw

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3 years ago